A uniform thin rod AB of length L has linear mass density `mu(x) = a + (bx)/(L)`, where x is measured from A. If the CM of the rod lies at a distance of `((7)/(12)L)` from A, then a and b are related as :_

Options

A. `2a = b `
B. `a = 2b `
C. `a = b `
D. `3a = 2b`

Answer

To solve the problem, we need to find the relationship between the constants \( a \) and \( b \) given that the center of mass (CM) of the rod lies at a distance of \( \frac{7}{12}L \) from point A.

### Step-by-Step Solution:

1. **Define the linear mass density**:
The linear mass density of the rod is given by:
\[ \mu(x) = a + \frac{bx}{L} \]
where \( x \) is the distance from point A.

2. **Determine the mass of a small segment**:
Consider a small segment of the rod of length \( dx \) located at a distance \( x \) from A. The mass \( dm \) of this small segment is:
\[ dm = \mu(x) \, dx = \left(a + \frac{bx}{L}\right) dx \]

3. **Calculate the total mass of the rod**:
The total mass \( M \) of the rod can be found by integrating \( dm \) from \( 0 \) to \( L \):
\[ M = \int_0^L dm = \int_0^L \left(a + \frac{bx}{L}\right) dx \]
This can be split into two integrals:
\[ M = \int_0^L a \, dx + \int_0^L \frac{bx}{L} \, dx \]
Evaluating these integrals gives:
\[ M = aL + \frac{b}{L} \cdot \frac{L^2}{2} = aL + \frac{bL}{2} \]
Thus,
\[ M = aL + \frac{bL}{2} = L\left(a + \frac{b}{2}\right) \]

4. **Calculate the center of mass**:
The center of mass \( x_{cm} \) is given by:
\[ x_{cm} = \frac{1}{M} \int_0^L x \, dm \]
Substituting \( dm \):
\[ x_{cm} = \frac{1}{M} \int_0^L x \left(a + \frac{bx}{L}\right) dx \]
This can be expanded as:
\[ x_{cm} = \frac{1}{M} \left( \int_0^L ax \, dx + \int_0^L \frac{bx^2}{L} \, dx \right) \]
Evaluating these integrals:
\[ \int_0^L ax \, dx = a \cdot \frac{L^2}{2} \]
\[ \int_0^L \frac{bx^2}{L} \, dx = \frac{b}{L} \cdot \frac{L^3}{3} = \frac{bL^2}{3} \]
Thus,
\[ x_{cm} = \frac{1}{M} \left( a \cdot \frac{L^2}{2} + \frac{bL^2}{3} \right) \]

5. **Substituting for \( M \)**:
Now substituting \( M \):
\[ x_{cm} = \frac{1}{L\left(a + \frac{b}{2}\right)} \left( \frac{aL^2}{2} + \frac{bL^2}{3} \right) \]
Simplifying this gives:
\[ x_{cm} = \frac{L}{\left(a + \frac{b}{2}\right)} \left( \frac{a}{2} + \frac{b}{3} \right) \]

6. **Setting \( x_{cm} \) equal to \( \frac{7}{12}L \)**:
Set the expression for \( x_{cm} \) equal to \( \frac{7}{12}L \):
\[ \frac{L}{\left(a + \frac{b}{2}\right)} \left( \frac{a}{2} + \frac{b}{3} \right) = \frac{7}{12}L \]
Dividing both sides by \( L \) (assuming \( L \neq 0 \)):
\[ \frac{\frac{a}{2} + \frac{b}{3}}{a + \frac{b}{2}} = \frac{7}{12} \]

7. **Cross-multiplying and simplifying**:
Cross-multiplying gives:
\[ 12\left(\frac{a}{2} + \frac{b}{3}\right) = 7\left(a + \frac{b}{2}\right) \]
Expanding both sides:
\[ 6a + 4b = 7a + \frac{7b}{2} \]
Rearranging terms:
\[ 6a + 4b - 7a - \frac{7b}{2} = 0 \]
This simplifies to:
\[ -a + 4b - \frac{7b}{2} = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ -2a + 8b - 7b = 0 \implies -2a + b = 0 \implies b = 2a \]

### Final Result:
Thus, the relationship between \( a \) and \( b \) is:
\[ b = 2a \]